What quantity of heat (in kJ) will be absorbed by a 65.5 g piece of aluminum (specific heat = 0.930 J/g °C) as it changes temperature from 23.0 °C to 67.0 °C? I kJ 1 2 3 4 5 6 7 8 9 +/- 0 C x 10
What quantity of heat (in kJ) will be absorbed by a 65.5 g piece of aluminum (specific heat = 0.930 J/g °C) as it changes temperature from 23.0 °C to 67.0 °C? I kJ 1 2 3 4 5 6 7 8 9 +/- 0 C x 10
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Question 20 of 27**
**Physics: Heat Transfer**
**Problem Statement:**
What quantity of heat (in kJ) will be absorbed by a 65.5 g piece of aluminum (specific heat = 0.930 J/g·°C) as it changes temperature from 23.0 °C to 67.0 °C?
**Calculator Input:**
A calculator is visible at the bottom of the screen with the following keys:
- Numerals (0-9)
- Basic arithmetic operations (+, -, x, ÷)
- Decimal point (.)
- Plus/Minus (±)
- "x10" for scientific notation
- Clear (C)
- Equals (=)
**Explanation:**
To solve this problem, the following formula for heat transfer can be used:
\[ Q = m \times c \times \Delta T \]
Where:
- \( Q \) is the quantity of heat (in Joules, J).
- \( m \) is the mass of the substance (in grams, g).
- \( c \) is the specific heat capacity (in J/g·°C).
- \( \Delta T \) is the change in temperature (in °C).
Given data:
- Mass (\( m \)) = 65.5 g
- Specific heat capacity (\( c \)) = 0.930 J/g·°C
- Initial temperature (\( T_i \)) = 23.0 °C
- Final temperature (\( T_f \)) = 67.0 °C
Step-by-Step Calculation:
1. Calculate the temperature change (\( \Delta T \)):
\[ \Delta T = T_f - T_i = 67.0 °C - 23.0 °C = 44.0 °C \]
2. Calculate the quantity of heat (\( Q \)):
\[ Q = 65.5 \, \text{g} \times 0.930 \, \frac{\text{J}}{\text{g·°C}} \times 44.0 \, \text{°C} \]
\[ Q = 2652.18 \, \text{J} \]
3. Convert the heat (\( Q \)) into kilojoules (kJ) because 1 kJ = 1000 J:
\[ Q = \frac{2652.18](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6895f9cd-8521-4c9a-bc62-df11c4ac3cae%2Fa344e9ca-10ef-4576-8668-4bea4b481420%2F4v2qrd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 20 of 27**
**Physics: Heat Transfer**
**Problem Statement:**
What quantity of heat (in kJ) will be absorbed by a 65.5 g piece of aluminum (specific heat = 0.930 J/g·°C) as it changes temperature from 23.0 °C to 67.0 °C?
**Calculator Input:**
A calculator is visible at the bottom of the screen with the following keys:
- Numerals (0-9)
- Basic arithmetic operations (+, -, x, ÷)
- Decimal point (.)
- Plus/Minus (±)
- "x10" for scientific notation
- Clear (C)
- Equals (=)
**Explanation:**
To solve this problem, the following formula for heat transfer can be used:
\[ Q = m \times c \times \Delta T \]
Where:
- \( Q \) is the quantity of heat (in Joules, J).
- \( m \) is the mass of the substance (in grams, g).
- \( c \) is the specific heat capacity (in J/g·°C).
- \( \Delta T \) is the change in temperature (in °C).
Given data:
- Mass (\( m \)) = 65.5 g
- Specific heat capacity (\( c \)) = 0.930 J/g·°C
- Initial temperature (\( T_i \)) = 23.0 °C
- Final temperature (\( T_f \)) = 67.0 °C
Step-by-Step Calculation:
1. Calculate the temperature change (\( \Delta T \)):
\[ \Delta T = T_f - T_i = 67.0 °C - 23.0 °C = 44.0 °C \]
2. Calculate the quantity of heat (\( Q \)):
\[ Q = 65.5 \, \text{g} \times 0.930 \, \frac{\text{J}}{\text{g·°C}} \times 44.0 \, \text{°C} \]
\[ Q = 2652.18 \, \text{J} \]
3. Convert the heat (\( Q \)) into kilojoules (kJ) because 1 kJ = 1000 J:
\[ Q = \frac{2652.18
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